Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 10, Problem 10.120QE
Interpretation Introduction

Interpretation:

Lewis structure for ethylene oxide, ethylene glycol, and acrylonitrile has to be drawn and also the hybrid orbitals of the central atom have to be given along with the number of pi bonds present.

Concept Introduction:

Lewis structure is used for predicting the shape of molecules.  From the steric number obtained in a Lewis structure, the molecular geometry can be predicted.  VSEPR model can predict the shape of molecules considering their Lewis structure.  Certain rules has to be followed in for the VSEPR model.

  • The molecule will have a shape where there is minimal electrostatic repulsion between the valence‑shell electron pairs.
  • The forces of repulsion between two lone pairs of electrons will be higher than the repulsion between lone pair and bond pair of electrons.  This in turn will be higher than the bond pair‑bond pair of electrons.

The hybridized orbitals and the steric number can be related as shown below;

Steric numberHybridized orbital
2sp
3sp2
4sp3
5sp3d
6sp3d2

Expert Solution & Answer
Check Mark

Explanation of Solution

Ethylene oxide:

Formula for ethylene oxide is C2H4O.

The total number of valence electrons is calculated as shown below;

    TotalNoofvalenceelectrons=(2×4)+(4×1)+(1×6)=8+4+6=18

A total of 14 electrons are involved in the skeletal structure.  Two lone pair of electrons are placed over the oxygen atom.  Hence, the Lewis structure of ethylene oxide is given as shown below;

Chemistry: Principles and Practice, Chapter 10, Problem 10.120QE , additional homework tip  1

Hybrid orbitals of central carbon atoms:

The carbon atom has does not have a lone pair of electrons and it is bonded to four atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=0+4=4

As the steric number is four, the hybridization of carbon atom is sp3.

Hybrid orbitals of central oxygen atom:

The oxygen atom has two lone pair of electrons and it is bonded to two atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=2+2=4

As the steric number is four, the hybridization of oxygen atom is sp3.

There is no unhybridized orbital left out for the formation of pi bonds.  Hence, there is no pi bonds in ethylene oxide.

Ethylene glycol:

Formula for ethylene glycol is HOCH2CH2OH.

The total number of valence electrons is calculated as shown below;

    TotalNoofvalenceelectrons=(2×4)+(6×1)+(2×6)=8+6+12=26

A total of 18 electrons are involved in the skeletal structure.  Two lone pair of electrons are placed over each oxygen atom.  Hence, the Lewis structure of ethylene glycol is given as shown below;

Chemistry: Principles and Practice, Chapter 10, Problem 10.120QE , additional homework tip  2

Hybrid orbitals of central carbon atoms:

The carbon atom has does not have a lone pair of electrons and it is bonded to four atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=0+4=4

As the steric number is four, the hybridization of carbon atom is sp3.

Hybrid orbitals of central oxygen atoms:

The oxygen atom has two lone pair of electrons and it is bonded to two atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=2+2=4

As the steric number is four, the hybridization of oxygen atom is sp3.

There is no unhybridized orbital left out for the formation of pi bonds.  Hence, there is no pi bonds in ethylene glycol.

Acrylonitrile:

Formula for acrylonitrile is CH2CHCN.

The total number of valence electrons is calculated as shown below;

    TotalNoofvalenceelectrons=(3×4)+(3×1)+(1×5)=12+3+5=20

A total of 12 electrons are involved in the skeletal structure.  The Lewis structure of acrylonitrile is given as shown below;

Chemistry: Principles and Practice, Chapter 10, Problem 10.120QE , additional homework tip  3

Hybrid orbitals of carbon atom C-1:

The carbon atom has does not have a lone pair of electrons and it is bonded to three atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=0+3=3

As the steric number is three, the hybridization of carbon atom is sp2.

Hybrid orbitals of carbon atom C-2:

The carbon atom has does not have a lone pair of electrons and it is bonded to three atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=0+3=3

As the steric number is three, the hybridization of carbon atom is sp2.

Hybrid orbitals of carbon atom C-3:

The carbon atom has does not have a lone pair of electrons and it is bonded to two atoms.  Therefore, the steric number is calculated as shown below;

    Stericnumber=0+2=2

As the steric number is two, the hybridization of carbon atom is sp.

There are one double bond and one triple bond present in the structure.  As multiple bonds are present, acrylonitrile contains pi bonds.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Phosphorus pentachloride, a key industrial compound with annual world production of about 2 x 107 kg, is used to make other compounds. It reacts with sulfur dioxide to produce phosphorus oxychloride (POCl3) and thionyl chloride (SOCl2). Draw a Lewis structure, and name the molecular shape of each product.
Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H2SO4, which has twooxygen atoms and two OH groups bonded to the sulfur.
Consider peroxynitrite (chemical formula: ONOO−), a structural isomer of the nitrate anion. It is generated in the cell when nitric oxide (NO) reacts with the superoxide radical anion (O2-.). Peroxynitrite is a potent oxidant and nitrating agent and can lead to DNA and protein damage (this is the complete question) Draw the correct Lewis structure for peroxynitrite and indicate approximate bond angles. Tell how many molecular degrees of freedom of motion are present in peroxynitrite and what motions they correspond to.

Chapter 10 Solutions

Chemistry: Principles and Practice

Ch. 10 - Which atomic orbitals overlap to form the bonds in...Ch. 10 - Prob. 10.12QECh. 10 - Identify the hybrid orbitals used by boron in BCl3...Ch. 10 - Identify the hybrid orbitals used by antimony in...Ch. 10 - Prob. 10.15QECh. 10 - Prob. 10.16QECh. 10 - Prob. 10.17QECh. 10 - Prob. 10.18QECh. 10 - Prob. 10.19QECh. 10 - Prob. 10.20QECh. 10 - Compare and contrast the molecular orbital and...Ch. 10 - Describe the bonding in molecular orbital terms...Ch. 10 - Prob. 10.23QECh. 10 - Prob. 10.24QECh. 10 - Prob. 10.25QECh. 10 - Prob. 10.26QECh. 10 - Prob. 10.27QECh. 10 - Prob. 10.28QECh. 10 - Prob. 10.29QECh. 10 - Prob. 10.30QECh. 10 - Prob. 10.31QECh. 10 - Prob. 10.32QECh. 10 - Prob. 10.33QECh. 10 - Prob. 10.34QECh. 10 - Prob. 10.35QECh. 10 - Prob. 10.36QECh. 10 - Prob. 10.37QECh. 10 - Prob. 10.38QECh. 10 - Prob. 10.39QECh. 10 - Use the VSEPR model to predict the bond angles...Ch. 10 - Prob. 10.41QECh. 10 - Prob. 10.42QECh. 10 - For each of the following molecules, complete the...Ch. 10 - Prob. 10.44QECh. 10 - Prob. 10.45QECh. 10 - Prob. 10.46QECh. 10 - Indicate which molecules are polar and which are...Ch. 10 - Prob. 10.48QECh. 10 - Indicate which of the following molecules are...Ch. 10 - Prob. 10.50QECh. 10 - Prob. 10.51QECh. 10 - Prob. 10.52QECh. 10 - Prob. 10.53QECh. 10 - Prob. 10.54QECh. 10 - Prob. 10.55QECh. 10 - Prob. 10.56QECh. 10 - Prob. 10.57QECh. 10 - Prob. 10.58QECh. 10 - Prob. 10.59QECh. 10 - Prob. 10.60QECh. 10 - Prob. 10.61QECh. 10 - Prob. 10.62QECh. 10 - Prob. 10.63QECh. 10 - Prob. 10.64QECh. 10 - Prob. 10.65QECh. 10 - Prob. 10.66QECh. 10 - Prob. 10.67QECh. 10 - Prob. 10.68QECh. 10 - Prob. 10.69QECh. 10 - Prob. 10.70QECh. 10 - Prob. 10.71QECh. 10 - Prob. 10.72QECh. 10 - Identify the orbitals on each of the atoms that...Ch. 10 - Prob. 10.74QECh. 10 - Prob. 10.75QECh. 10 - How many sigma bonds and how many pi bonds are...Ch. 10 - Give the hybridization of each central atom in the...Ch. 10 - Prob. 10.78QECh. 10 - Prob. 10.79QECh. 10 - Prob. 10.80QECh. 10 - Prob. 10.81QECh. 10 - Predict the hybridization at each central atom in...Ch. 10 - Prob. 10.83QECh. 10 - Tetrafluoroethylene, C2F4, is used to produce...Ch. 10 - Prob. 10.85QECh. 10 - Prob. 10.86QECh. 10 - Prob. 10.87QECh. 10 - Prob. 10.88QECh. 10 - Prob. 10.89QECh. 10 - Prob. 10.90QECh. 10 - Prob. 10.91QECh. 10 - Prob. 10.92QECh. 10 - Prob. 10.93QECh. 10 - Prob. 10.94QECh. 10 - Prob. 10.95QECh. 10 - Prob. 10.96QECh. 10 - Prob. 10.97QECh. 10 - Prob. 10.98QECh. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - Prob. 10.102QECh. 10 - Prob. 10.103QECh. 10 - Prob. 10.104QECh. 10 - Prob. 10.105QECh. 10 - Following are the structures of three isomers of...Ch. 10 - The ions ClF2 and ClF2+ have both been observed....Ch. 10 - Aspirin, or acetylsalicylic acid, has the formula...Ch. 10 - Aspartame is a compound that is 200 times sweeter...Ch. 10 - Prob. 10.110QECh. 10 - Prob. 10.111QECh. 10 - Calcium cyanamide, CaNCN, is used both to kill...Ch. 10 - Histidine is an essential amino acid that the body...Ch. 10 - Formamide, HC(O)NH2, is prepared at high pressures...Ch. 10 - Prob. 10.115QECh. 10 - Prob. 10.116QECh. 10 - Prob. 10.117QECh. 10 - Prob. 10.118QECh. 10 - Prob. 10.119QECh. 10 - Prob. 10.120QECh. 10 - Prob. 10.121QECh. 10 - Prob. 10.122QECh. 10 - Prob. 10.123QECh. 10 - Prob. 10.124QECh. 10 - Two compounds have the formula S2F2. Disulfur...Ch. 10 - Prob. 10.126QECh. 10 - Prob. 10.127QE
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY